Class 9 Chapter 9 (Areas of parallelogram and triangles) Notes

Notes of Chapter 9 Area of Parallelograms and Triangles

Topics in the Chapter

• Area of Plane Figures
• Fundamentals
• Theorems

Areas of Plane Figures

• Two congruent figures have equal areas.
• A diagonal of a parallelogram divides it into two triangles of equal area.
• Parallelograms on the same base (or equal base) and between the same parallels are equal in area.
• Triangles on the same base (or equal bases) and between the same parallels are equal in area.
• Area of a parallelogram = base × height.
• Area of a triangle = 1/2 ×base × height
• A median of a triangle divides it into two triangles of equal area.
• Diagonals of a parallelogram divides it into four triangles of equal area.

Fundamentals

• Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 

Examples:

 

 

• Area of triangle is half the product of its base (or any side) and the corresponding altitude (or height).
• Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
• A median of a triangle divides it into two triangles of equal areas. 

Theorems:

(1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.

Given:  A parallelogram ABCD in which BD is one of the diagonals.

To prove:   

Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that it is sufficient to show that

ΔABD ≅ ΔCDB

In Δs ABD and CDB, we have

AB = CD

AD = CB

And, BD = DB

So, by SSS criterion of congruence, we have

ΔABD ≅ ΔCDB

Hence, ar(ΔABD) = ar(ΔCDB)

(2) Prove that parallelograms on the same base and between the same parallels are equal in area.

Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.

To prove: ar(||gm ABCD) = ar(||gm ABCD)

Proof: In Δs ADF and BCE, we have

AD = BC

AF = BE

And, ∠DAF = ∠CBE         [ ⸪ AD ∥ BC and AF ∥ BE]

So, by SAS criterion of congruence, we have

ΔADF ≅ ΔBCE

ar(ΔADF) = ar(ΔBCE)   …..(i)

Now, ar(||gm ABCD) = ar(square ABED) + ar(ΔBCE)

ar(||gm ABCD) = ar(square ABED) + ar(ΔADF)    [Using(i)]

ar(||gm ABCD) = ar(||gm ABEF)

Hence, ar(||gm ABCD) = ar(||gm ABEF)

(3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.

Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.

To prove: ar(parallelogram ABCD) = A B × AL

Construction: Complete the rectangle ALMB by drawing BM⊥CD.

Proof: Since ar(parallelogram ABCD) and rectangle ALMB are on the same base and between the same parallels.

ar(parallelogram ABCD)

= ar(rect.ALMB)

=AB × AL    [By rect. Area axiom area of a rectangle = Base \(\times\) Height]

Hence, ar(parallelogram ABCD) = AB×AL

(4) Prove that parallelograms on equal bases and between the same parallels are equal in area.

Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.

To prove: ar(parallelogram ABCD) = ar(parallelogram PQRS)

Construction: Draw AL ⊥ DR and PM ⊥ DR

Proof: Since AB ⊥ DR, AL ⊥ DR and PM ⊥ DR

AL = PM

Now, ar(parallelogram ABCD) = AB × AL

ar(parallelogram ABCD) = PQ × PM  [AB = PQ and AL = PM]

ar(parallelogram ABCD) = ar(parallelogram PQRS)

(5) Prove that triangles on the same bases and between the same parallels are equal in area.

Proof: We have,

BD∥CA

And, BC∥DA

sq.BCAD is a parallelogram.

Similarly, sq.BCQP is a parallelogram.

Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.

ar(parallelogram BCQP) = ar(parallelogram BCAD)    ….(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

ar(ΔPBC) = \(\dfrac{1}{2}\)ar(parallelogram BCQP)   …..(ii)

And, ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCAQ)  ....(iii)

Now, ar(parallelogram BCQP) = ar(parallelogram BCAD) [ From (i)]

\(\dfrac{1}{2}\)ar(parallelogram BCQP) = \(\dfrac{1}{2}\)ar(parallelogram BCAD)

ar(ΔABC) = ar(ΔPBC)    [From (ii) and (iii)]

Hence, ar(ΔABC) = ar(ΔPBC)

(6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.

Given: A ΔABC in which AL is the altitude to the side BC.

To prove:

Construction: Through C and A draw CD ∥ BA and AD ∥ BC respectively, intersecting each other at D.

Proof: We have,

BA∥CD and AD ∥ BC

So, BCDA is a parallelogram.

Since AC is a diagonal of parallelogram BCDA.

[BC is the base and AL is the corresponding altitude of parallelogram BCDA]

(7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.

Given: A ΔABC and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.

To prove: ar(ΔABC) = \(\dfrac{1}{2}\)ar(||gm BCDE)

Construction: Draw AL ⊥ BC and DM ⊥ BC, meeting BC produced in M.

Proof: Since A, E and D are collinear and BC ∥ AD

AL = DM    …..(i)

Now,

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × AL)

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × DM)   [AL = DM (from (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCDE)

 

(8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.

Given: A trapezium ABCD in which AB ∥ DC; AB = a, DC = b and AL = CM = h, where AL⊥DC and CM ⊥ AB.

To prove: ar(trap. ABCD) = \(\dfrac{1}{2}\)h × (a + b)

Construction: Join AC

Proof: We have,

ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)

ar(trap. ABCD) = \(\dfrac{1}{2}\)(AB × CM) + \(\dfrac{1}{2}\)(DC × AL)

ar(trap. ABCD) = \(\dfrac{1}{2}\)ah + \(\dfrac{1}{2}\)bh  [AB = a and DC = b]

ar(trap. ABCD) = \(\dfrac{1}{2}\)h×(a+b)

(9) Prove that triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal.

Given: Two triangles ABC and PQR such that:

ar(ΔABC) = ar(ΔPQR)

AB = PQ

CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.

To prove: CN = RT

Proof: In ΔABC, CN is the altitude corresponding to side AB.

ar(ΔABC) = \(\dfrac{1}{2}\)(AB × CN)    ….(i)

Similarly, we have,

ar(ΔPQR) = \(\dfrac{1}{2}\)(PQ × RT)   …..(ii)

Now, ar(ΔABC) = ar(ΔPQR)

\(\dfrac{1}{2}\)(AB × CN) = \(\dfrac{1}{2}\)(PQ × RT)

(AB × CN) = (PQ × RT)

(PQ × CN) = (PQ × RT)    [ AB = PQ (Given)]

CN = RT

(10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD are such that

ar(ΔABD) = ar(ΔCDB) and ar(ΔABC) = ar(ΔACD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,

ar(ΔABC) = ar(ΔACD)    …..(i)

But, ar(ΔABC) + ar(ΔACD) = ar(ABCD)

2ar(ΔABC) = ar(ABCD)    [Using (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(ABCD) ….(ii)

Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.

ar(ΔABD) = ar(ΔBCD)   ….(iii)

But, ar(ΔABD) + ar(ΔBCD) = ar(ABCD)

2ar(ΔABD) = ar(ABCD)    [Using(iii)]

ar(ΔABD) = \(\dfrac{1}{2}\)ar(ABCD)   …..(iv)

From (ii) and (iv), we get

ar(ΔABC) = ar(ΔABD)

Since Δs ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.

i.e. Altitude from C of ΔABC = Altitude from D of ΔABD

DC∥AB

Similarly, AD∥BC

Hence, quadrilateral ABCD is a parallelogram.

(11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.

Given: A rhombus ABCD whose diagonals AC and BD intersect at O.

To prove: ar(rhombus ABCD) = \(\dfrac{1}{2}\)(AC×BD)

Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,

OB ⊥ AC and OD ⊥ AC

ar(rhombus) = ar(ΔABC) + ar(ΔADC)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × BO) + \(\dfrac{1}{2}\)(AC × DO)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × (BO + DO))

ar(rhombus) =\(\dfrac{1}{2}\)(AC×BD)

(12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.

Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.

To prove: ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.

OA = OC and OB = OD

Also, the median of a triangle divides it into two equal parts.

Now, in ΔABC, BO is the median.

ar(ΔOAB) = ar(ΔOBC)  ….(i)

In ΔBCD, CO is the median

ar(ΔOBC) = ar(ΔOCD)    …..(ii)

In ΔACD, DO is the median

ar(ΔOCD) = ar(ΔAOD)   ….(iii)

From (i), (ii) and (iii), we get

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

(13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: We have,

ar(ΔAOD) = ar(ΔBOC)

ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC) + ar(ΔAOB)

ar(ΔABD) = ar(ΔABC)

Thus, Δs ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.

Altitude from ΔABD Altitude from C of ΔABC

DC ∥ AB

Similarly, we have, AD ∥ BC.

Hence, quadrilateral ABCD is a parallelogram.

(14) Prove that a median of a triangle divides it into two triangles of equal area.

Given: A ΔABC in which AD is the median.

To prove: ar(ΔABD) = ar(ΔADC)

Construction: Draw AL ⊥ BC.

Proof: Since AD is the median of ΔABC. Therefore, D is the mid point of BC.

BD = DC

BD × AL = DC × AL  [ Multiplying both sides by AL]

\(\dfrac{1}{2}\)(BD × AL) = \(\dfrac{1}{2}\)(DC × AL)

ar(ΔABD) = ar(ΔADC)

ALITER, Since Δs ABD and ADC have equal bases and the same altitude AL.

ar(ΔABD) = ar(ΔADC)

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